---
title: "NCERT Solutions Class 9 Science Exploring Mixtures and their Separation"
url: https://www.swavid.com/science/class/9/chapter/exploring-mixtures-and-their-separation/ncert-solutions
dateModified: 2026-10-07T16:27:34+00:00
---

# NCERT Solutions Class 9 Science Exploring Mixtures and their Separation

This chapter's questions cover the classification of mixtures, methods of separation such as crystallization, distillation, chromatography, and the properties of solutions, suspensions, and colloids.

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## Think It Over

### Question 1

*3 marks · Short answer*

Why do suspended particles settle in muddy water over time but not in milk?

**Solution**

1. Muddy water is a suspension where large insoluble solid particles are suspended in water and settle down over time due to gravity.
2. Milk is a colloidal mixture in which the dispersed particles are much smaller in size.
3. Due to continuous thermal movements and uniform dispersion, colloidal particles in milk do not settle over time.

**Answer:** Suspended particles in muddy water are large and heavy, causing them to settle over time, whereas particles in milk are colloidal and remain uniformly dispersed without settling.

> Common mistake: Thinking that milk is a true solution instead of a colloid.

### Question 2

*3 marks · Short answer*

How is evaporation different from boiling?

**Solution**

1. Evaporation is a surface phenomenon where a liquid changes into vapours at any temperature below its boiling point.
2. Boiling is a bulk phenomenon where a liquid changes into vapours rapidly throughout the entire liquid at a fixed boiling point.
3. Evaporation causes cooling, whereas boiling requires a continuous supply of heat energy.

**Answer:** Evaporation occurs slowly at any temperature at the surface of the liquid, while boiling occurs rapidly throughout the liquid at a specific boiling point.

> Common mistake: Confusing evaporation as occurring only at high temperatures.

### Question 3

*3 marks · Short answer*

Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?

**Solution**

1. Air contains tiny dust and smoke particles which form a colloidal mixture or suspension with gases.
2. When sunlight passes through small gaps between the leaves, it hits these suspended particles in the air.
3. The particles scatter the light in different directions, making the path of the light beam visible, which is known as the Tyndall effect.

**Answer:** Bright rays of sunlight are seen due to the Tyndall effect, where dust and smoke particles in the air scatter the light passing through gaps in the leaves.

> Common mistake: Stating that air reflects light instead of scattering it.

## Activity 5.1

### Question 4

*Activity*

Are the particles visible in each mixture? Record your observations.

**Solution**

1. In beaker A (salt and water), the solute particles are not visible as it forms a homogeneous solution.
2. In beaker B (chalk powder and water), the chalk particles are clearly visible to the naked eye as it forms a suspension.
3. In beaker C (milk and water), the particles are not individually visible to the naked eye as it forms a colloidal mixture.

**Answer:** Particles are visible only in the suspension (beaker B) and not in the solution (beaker A) or colloid (beaker C).

> Common mistake: Stating that milk particles are visible to the naked eye.

### Question 5

*Activity*

Direct the light from a laser pointer through the beakers containing the mixtures (Fig. 5.3) and observe it from the side of the beaker in a direction perpendicular to the laser beam. Record your observations.

**Solution**

1. In beaker A containing salt and water (solution), the path of the laser beam is not visible because solute particles are too small to scatter light.
2. In beaker B containing chalk powder and water (suspension), the path of the laser beam is scattered and visible.
3. In beaker C containing milk and water (colloid), the path of the laser beam is clearly visible due to the scattering of light by colloidal particles.

**Answer:** The path of the laser beam is visible in the suspension and colloid, but not in the true solution.

> Common mistake: Confusing solutions with colloids regarding the Tyndall effect.

### Question 6

*Activity*

Predict what you would observe in each of the beakers if you leave them undisturbed for a few minutes.

**Solution**

1. In beaker A (salt and water), the salt remains dissolved and does not settle down over time.
2. In beaker B (chalk powder and water), the heavier chalk particles slowly settle down at the bottom of the beaker after some time.
3. In beaker C (milk and water), the particles remain uniformly dispersed and do not settle down over time.

**Answer:** Chalk powder settles down in the suspension, whereas salt solution and milk remain stable without settling.

> Common mistake: Assuming colloidal particles settle down like suspension particles over time.

### Question 7

*Activity*

Set up a filtration apparatus and filter each mixture separately. Is there any residue left on the filter paper?

**Solution**

1. When filtering beaker A (salt and water), no residue is left on the filter paper as salt is completely dissolved.
2. When filtering beaker B (chalk powder and water), chalk powder is left behind as a residue on the filter paper.
3. When filtering beaker C (milk and water), no residue is left on the filter paper as colloidal particles pass through it.

**Answer:** Residue is left on the filter paper only for the suspension (beaker B).

> Common mistake: Believing that colloidal particles like milk can be separated by ordinary filter paper.

### Question 8

*Activity*

Based on your observations, do you think these are the same types of mixtures or are they different?

**Solution**

1. Based on particle visibility, Tyndall effect, settling, and filtration behavior, the three mixtures are fundamentally different.
2. Beaker A forms a homogeneous mixture or solution.
3. Beaker B forms a heterogeneous mixture or suspension.
4. Beaker C forms a colloidal mixture.

**Answer:** They are different types of mixtures: a solution, a suspension, and a colloid.

> Common mistake: Treating milk and sugar solution as the same type of mixture.

## Pause and Ponder

### Question 1

*3 marks · Numerical*

A common talcum powder contains $4\text{ }\%\text{ m/m}$ zinc oxide, which acts as an antiseptic. How much zinc oxide is present in $300\text{ g}$ of the talcum powder?

**Solution**

1. Mass by mass percentage of zinc oxide = $4\text{ }\%\text{ m/m}$
2. Total mass of talcum powder = $300\text{ g}$
3. Mass of zinc oxide = $\frac{\text{Mass by mass percentage} \times \text{Total mass of solution}}{100}$
4. Mass of zinc oxide = $\frac{4}{100} \times 300\text{ g} = 12\text{ g}$

**Answer:** $12\text{ g}$

> Common mistake: Dividing by 300 instead of multiplying by the total mass.

### Question 2

*3 marks · Numerical*

Your mother gives you a bottle of orange juice concentrate to mix with water and serve it to your visiting friends. She asks you to mix two tablespoons of the concentrate with water in a glass tumbler. If each tablespoon measures $15\text{ mL}$ and you make $150\text{ mL}$ of juice per person, what is the $\%\text{ v/v}$ of orange juice concentrate in the mixture you prepared?

**Solution**

1. Given: Volume of orange juice concentrate (solute) = $2 \times 15\text{ mL} = 30\text{ mL}$
2. Given: Total volume of solution = $150\text{ mL}$
3. Formula: $\text{Volume by volume percentage} = \frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100$
4. Substitution: $\text{Volume by volume percentage} = \frac{30\text{ mL}}{150\text{ mL}} \times 100$
5. Result: $20\%\text{ v/v}$

**Answer:** $20\%\text{ v/v}$

> Common mistake: Forgetting to multiply the number of tablespoons by the volume of each tablespoon.

### Question 3

*3 marks · Short answer*

Vinegar, used as a food preservative and additive, contains $5\%\text{ v/v}$ acetic acid. Glacial acetic acid is a liquid, i.e., $100\%$ acetic acid. If you want to make vinegar from glacial acetic acid, how would you proceed?

**Solution**

1. A $5\%\text{ v/v}$ acetic acid solution (vinegar) means that $5\text{ mL}$ of pure acetic acid is present in $100\text{ mL}$ of the total solution.
2. To prepare this, take $5\text{ mL}$ of glacial acetic acid ($100\%$ acetic acid).
3. Add sufficient water to glacial acetic acid to make the total volume of the solution exactly $100\text{ mL}$ and stir well.

**Answer:** Take $5\text{ mL}$ of glacial acetic acid and add water to make the total volume $100\text{ mL}$.

> Common mistake: Adding $5\text{ mL}$ of acid to $100\text{ mL}$ of water instead of making the total volume $100\text{ mL}$.

## Activity 5.2

### Question 1

*3 marks · Short answer*

Based on the information from the above graph, predict which of the two compounds, 'A' or 'B', will dissolve more in a given amount of water at a given temperature?

**Solution**

1. Examine the solubility curves for compounds 'A' and 'B' in Fig. 5.6.
2. At any given temperature on the x-axis, read the corresponding solubility on the y-axis (measured in grams per 100 g of water).
3. The curve for compound 'B' lies above the curve for compound 'A' across the temperature range, showing that compound 'B' dissolves more in 100 g of water than compound 'A'.

**Answer:** Compound 'B' will dissolve more than compound 'A' in a given amount of water at a given temperature.

> Common mistake: Confusing the two curves or misreading the y-axis values.

### Question 2

*3 marks · Fill in the blank*

Observe Fig. 5.6 and fill in the blanks of the following statements:
(i) The solubility of compound 'A' in water at $20\text{ }^\circ\text{C}$ is \_
(ii) The solubility of compound 'B' at $20\text{ }^\circ\text{C}$ is \_
(iii) The solubility of \_ increases more than that of \_ with an increase in the temperature.

**Part (i)**

1. Observe the solubility curve for compound 'A' in Fig. 5.6.
2. The solubility of compound 'A' at $20\text{ }^\circ\text{C}$ is lower than its solubility at $60\text{ }^\circ\text{C}$.

Answer (i): less than

**Part (ii)**

1. Observe the solubility curve for compound 'B' in Fig. 5.6.
2. The solubility of compound 'B' at $20\text{ }^\circ\text{C}$ is lower than its solubility at $60\text{ }^\circ\text{C}$.

Answer (ii): less than

**Part (iii)**

1. Compare the steepness of the curves for compounds 'A' and 'B' in Fig. 5.6.
2. The solubility curve for compound 'B' rises much more steeply than that of compound 'A' with an increase in temperature.

Answer (iii): compound 'B', compound 'A'

**Answer:** (i) less than, (ii) less than, (iii) compound 'B', compound 'A'

> Common mistake: Confusing the steepness of the curves or misreading the temperatures on the x-axis.

## Pause and Ponder

### Question 4

*3 marks · Short answer*

Refer to the solubility curves given in Activity 5.2. If equal masses of hot, saturated solutions of compounds 'A' and 'B' are cooled from $80\text{ }^\circ\text{C}$ to $60\text{ }^\circ\text{C}$, which solution is likely to deposit more solid?

**Solution**

1. From the solubility curves in Fig. 5.6, the solubility of compound 'B' decreases significantly when cooled from $80\text{ }^\circ\text{C}$ to $60\text{ }^\circ\text{C}$, whereas the solubility of compound 'A' changes very little over the same temperature range.
2. A larger decrease in solubility upon cooling means that more solute exceeds the saturation limit and separates out as solid crystals.
3. Therefore, the hot saturated solution of compound 'B' is likely to deposit more solid than compound 'A'.

**Answer:** Compound 'B' will deposit more solid.

> Common mistake: Confusing solubility with the absolute amount of solute without looking at the slope of the curve.

### Question 5

*3 marks · Short answer*

Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.

**Solution**

1. Yes, the rate of evaporation affects the size of common salt crystals formed during crystallization.
2. If the rate of evaporation is increased (such as by rapid heating), smaller and less well-formed crystals are obtained because particles get less time to arrange themselves in a regular geometric pattern.
3. If the rate of evaporation is decreased (slow evaporation), larger, well-shaped and shiny crystals are formed as particles get sufficient time to come together orderly.

**Answer:** Increasing the rate of evaporation produces smaller crystals, while decreasing it produces larger crystals.

> Common mistake: Thinking that fast evaporation leads to bigger crystals because more salt precipitates quickly.

## Activity 5.4

### Question 1

*3 marks · Short answer*

Observe Fig. 5.9, it shows how salt crystals are obtained from seawater. Can you describe the process in your own words?

**Solution**

1. Seawater is collected in shallow pools or lagoons and allowed to evaporate under the heat of the sun, leaving behind concentrated sea brine.
2. The concentrated solution undergoes further evaporation or boiling to form a saturated solution.
3. On cooling or further evaporation, salt crystals separate out from the saturated solution.

**Answer:** Seawater is first concentrated by solar evaporation to form a saturated solution, from which salt crystals are obtained.

> Common mistake: Confusing crystallization directly with simple boiling without mentioning the formation of a saturated solution first.

## Pause and Ponder

### Question 6

*4 marks · True or false*

State whether the following statements are True or False. Also, correct the False statements.
(i) Salt can be separated from a salt solution by evaporation or distillation.
(ii) Distillation can be used for separation of two liquids even when these have the same boiling point.
(iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.
(iv) Evaporation and crystallization are the same processes.

**Part (i)**

1. Evaporation allows the solvent to escape while leaving the solid salt behind.
2. Distillation allows both the recovery of the solvent and the separation of salt.
3. Therefore, the statement is true.

Answer (i): True

**Part (ii)**

1. Distillation requires liquids to have a sufficient difference in their boiling points (at least about $25\ ^\circ\text{C}$).
2. Liquids with the same boiling point vaporize together and cannot be separated by simple distillation.
3. Therefore, the statement is false.

Answer (ii): False. Distillation is used for the separation of two liquids that differ in boiling point by at least about $25\ ^\circ\text{C}$.

**Part (iii)**

1. In paper chromatography, the sample spot must remain above the solvent level so it does not dissolve directly into the solvent container.
2. Therefore, the statement is false.

Answer (iii): False. In paper chromatography, the solvent level should be below the sample spot at the beginning of the experiment.

**Part (iv)**

1. Evaporation involves removing the liquid completely to leave the solute behind, often resulting in small or unformed crystals.
2. Crystallization is a controlled process of forming large, well-shaped pure crystals from a saturated solution upon slow cooling.
3. Therefore, the statement is false.

Answer (iv): False. Evaporation removes the solvent completely, whereas crystallization produces pure solid crystals by cooling a saturated solution slowly.

**Answer:** Statements (i) is True, (ii) is False, (iii) is False, and (iv) is False.

> Common mistake: Confusing the initial position of the sample spot relative to the solvent level in chromatography.

## Pause and Ponder

### Question 7

*3 marks · Short answer*

Why do immiscible liquids form two separate layers in a separating funnel?

**Solution**

1. Immiscible liquids do not dissolve in each other and have different densities.
2. When placed in a separating funnel, the liquid with lower density forms the upper layer.
3. The liquid with higher density forms the lower layer due to gravity, creating two distinct layers.

**Answer:** Immiscible liquids form two separate layers because they have different densities and do not dissolve in each other.

> Common mistake: Thinking that boiling points or miscibility alone determines the layers instead of density differences.

### Question 8

*3 marks · Short answer*

Is sublimation different from evaporation? Justify.

**Solution**

1. Sublimation is the transition of a solid directly into a vapour (or gas) below its melting point without passing through the liquid state.
2. Evaporation is the phenomenon in which a liquid changes into a vapour at any temperature below its boiling point from the surface of the liquid.
3. Thus, sublimation involves a solid changing directly to a gas, whereas evaporation involves a liquid changing to a gas.

**Answer:** Yes, sublimation and evaporation are different processes; sublimation is the direct conversion of a solid to a vapour, while evaporation is the conversion of a liquid to a vapour below its boiling point.

> Common mistake: Confusing the initial states, thinking both involve liquid state or happen at the same temperatures.

## Pause and Ponder

### Question 9

*3 marks · Short answer*

Clouds are made up of tiny water droplets or ice crystals floating in the air. Based on what you know about solutions, suspensions and colloids, what type of mixture do you think clouds are and why?

**Solution**

1. Clouds consist of tiny water droplets or ice crystals suspended in air.
2. The particle size in clouds falls within the range of $1\text{ nm}$ to $1000\text{ nm}$, which is characteristic of colloids.
3. Therefore, clouds are colloidal mixtures (aerosols) where liquid or solid particles are dispersed in a gas.

**Answer:** Clouds are colloidal mixtures because their droplet sizes range between $1\text{ nm}$ and $1000\text{ nm}$ and they remain suspended in air without settling.

> Common mistake: Confusing clouds with suspensions just because water droplets are visible.

### Question 10

*3 marks · Short answer*

Why do cities with a lot of smoke and dust in the air often look hazy?

**Solution**

1. Smoke and dust particles present in polluted urban air form colloidal and suspension mixtures.
2. These suspended and colloidal particles scatter the sunlight passing through them.
3. This scattering of light, known as the Tyndall effect, makes the path of light visible and gives the atmosphere a hazy appearance.

**Answer:** Cities look hazy because smoke and dust particles scatter sunlight through the Tyndall effect.

> Common mistake: Stating that smoke dissolves in air instead of remaining suspended.

## Revise, Reflect, Refine

### Question 1

*1 mark · MCQ*

Which of the following mixtures are correctly classified as homogeneous (Hm) and heterogeneous (Ht)? Choose the correct option.

- Air — Hm, Milk — Ht, Sugar solution — Hm, Smoke — Hm
- Brass — Ht, Fog — Ht, Vinegar — Ht, Muddy water — Hm
- Copper sulfate solution — Hm, Salt solution — Hm, Milk — Hm, Bronze — Hm
- Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm

**Solution**

1. Air, sugar solution, and bronze (alloy) are homogeneous mixtures.
2. Muddy water, milk, and blood are heterogeneous mixtures (suspensions or colloids).
3. Option (iv) correctly classifies muddy water as heterogeneous, milk as heterogeneous, blood as heterogeneous, and brass as homogeneous.

**Answer:** (iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm

> Common mistake: Confusing colloids like milk and blood with homogeneous mixtures because they appear uniform.

### Question 2

*1 mark · MCQ*

Choose the correct options, and explain the reason for the correct and incorrect options. Which among the following mixtures show the Tyndall Effect? A mixture of: (a) air and dust particles (b) copper sulfate and water (c) starch and water (d) acetone and water

- a and b
- b and d
- a and c
- c and d

**Solution**

1. Air and dust particles form a heterogeneous mixture (colloid/suspension) that scatters light, thus showing the Tyndall effect.
2. Starch and water form a colloid/suspension that also shows the Tyndall effect.
3. Copper sulfate and water is a true solution, and acetone and water is a homogeneous solution; neither shows the Tyndall effect.

**Answer:** (iii) a and c

> Common mistake: Assuming all liquid mixtures exhibit the Tyndall effect without checking if they are solutions or colloids.

### Question 3

*3 marks · Short answer*

A mixture can be categorised as a solution, a suspension, or a colloid, each possessing distinct properties. Utilise the words or phrases provided in the box to fill in the Table 5.2. Words and phrases may be used more than once.

**Solution**

1. Solution: Nature is homogeneous, particle size is small ($< 1\text{ nm}$), transparent, does not settle down, cannot be separated by filtration, does not scatter light (no Tyndall effect). Examples: Salt solution, Brass.
2. Suspension: Nature is heterogeneous, particle size is large ($> 1000\text{ nm}$), settles down when left undisturbed, separates by filtration, scatters light. Examples: Mud, Sand in water.
3. Colloid: Nature is heterogeneous (appears homogeneous), particle size is moderate ($1-1000\text{ nm}$), does not settle down, cannot be separated by filtration, scatters light (Tyndall effect). Examples: Milk, Smoke, Butter.

**Answer:** Completed table for solution, suspension, and colloid with respective properties and examples.

> Common mistake: Mixing up particle size ranges or filtration properties between suspensions and colloids.

### Question 4

*3 marks · Numerical*

Solve the following problems:
(i) A cake recipe uses dry ingredients, namely $75\text{ g}$ of sugar for $420\text{ g}$ of all-purpose flour and $5\text{ g}$ of sodium hydrogencarbonate. Express the concentration of each component in the mixture using an appropriate method.
(ii) A brass alloy contains $70\%$ copper by mass. Calculate the quantities of copper and zinc present in $120\text{ g}$ of brass.

**Part (i)**

1. Total mass of the mixture = 75 g + 420 g + 5 g = 500 g.
2. Mass percentage of sugar = (75 g / 500 g) × 100 = 15%.
3. Mass percentage of flour = (420 g / 500 g) × 100 = 84%.
4. Mass percentage of sodium hydrogencarbonate = (5 g / 500 g) × 100 = 1%.

Answer (i): Sugar: 15%, Flour: 84%, Sodium hydrogencarbonate: 1%

**Part (ii)**

1. Mass of copper = 70% of 120 g = (70 / 100) × 120 g = 84 g.
2. Mass of zinc = Total mass - Mass of copper = 120 g - 84 g = 36 g.

Answer (ii): Copper: 84 g, Zinc: 36 g

**Answer:** The concentrations are sugar 15%, flour 84%, and sodium hydrogencarbonate 1%; the brass contains 84 g of copper and 36 g of zinc.

> Common mistake: Calculating the percentage based on the mass of the solvent instead of the total mass of the mixture.

### Question 5

*3 marks · Short answer*

The label on a cooking oil pack says one litre ($910\text{ g}$). If this oil is mixed with water, will it form a separate layer? If so, which substance will be on top? How will you separate the two layers? Also, draw the diagram of the apparatus used.

**Solution**

1. Since oil and water are immiscible and oil has a lower density ($910\text{ g}$ per litre, hence density $\approx 0.91\text{ g/cm}^3$) than water, they will form separate layers with cooking oil on top and water at the bottom.
2. The two layers can be separated using a separating funnel by opening the stopcock to run out the lower water layer and closing it when the oil reaches the stopcock.
3. Diagram: Draw a separating funnel supported on a laboratory stand containing an upper oil layer and a lower water layer, with a beaker placed below the stopcock as shown in Fig. 5.16 of the textbook.

**Answer:** Oil forms a separate layer on top of water and can be separated using a separating funnel.

> Common mistake: Confusing which liquid stays on top based on density.

### Question 6

*1 mark · Assertion and reason*

Assertion (A): Solutions do not exhibit the Tyndall effect.
Reason (R): The particles in solutions are larger than $100\text{ nm}$, so they cannot scatter light.
Choose the correct option:

- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true, but R is false.
- A is false, but R is true.

**Solution**

1. Solutions are homogeneous mixtures with particle sizes less than $1\text{ nm}$, which are too small to scatter light, so Assertion (A) is true.
2. Reason (R) states that solution particles are larger than $100\text{ nm}$, but actual solution particles are smaller than $1\text{ nm}$, making Reason (R) false.

**Answer:** (iii) A is true, but R is false.

> Common mistake: Confusing particle sizes of solutions with those of colloids or suspensions.

### Question 7

*3 marks · Short answer*

How would you separate the mixtures given in Table 5.3? Mention the reason for choosing your method. If a mixture cannot be separated, explain why.

**Part (i)**

1. Mud from muddy water: Filtration, as mud particles are suspended and larger than the filter paper pores.
2. Plasma from blood: Centrifugation, as it separates components based on density by spinning at high speed.
3. Naphthalene and sand: Sublimation, as naphthalene changes directly to vapour on heating while sand does not.
4. Chalk powder and common salt: Filtration, as chalk is insoluble in water while salt is soluble.
5. Common salt and water: Evaporation or Crystallization, as salt remains as a solid residue after the solvent evaporates.
6. Oil from water: Separating funnel, as they are immiscible liquids with different densities.
7. Pigments of the flower: Paper chromatography, as components move at different rates on the paper.

Answer (i): The methods are filtration, centrifugation, sublimation, filtration, evaporation/crystallization, separating funnel, and paper chromatography respectively.

**Answer:** The separation methods for the mixtures are: Mud from muddy water: Filtration; Plasma from blood: Centrifugation; Naphthalene and sand: Sublimation; Chalk powder and common salt: Filtration; Common salt and water: Evaporation or Crystallization; Oil from water: Separating funnel; Pigments of the flower: Paper chromatography.

> Common mistake: Confusing filtration with centrifugation for blood components.

### Question 8

*3 marks · Short answer*

Two miscible liquids, A and B, are present in a mixture. The boiling point of A is $60\text{ }^\circ\text{C}$ and the boiling point of B is $90\text{ }^\circ\text{C}$. Suggest a method to separate them. Also, draw a labelled diagram of the method suggested.

**Solution**

1. The mixture of liquids A and B can be separated by distillation because their boiling points differ by 30 °C, which is greater than the required 25 °C.
2. Diagram: Draw the distillation set-up as shown in Fig. 5.12, including the distillation flask, thermometer, water condenser, and receiving flask.
3. The liquid with the lower boiling point (A) vaporises first, passes through the condenser, and is collected as a pure liquid.

**Answer:** Distillation is the suitable method; refer to the distillation set-up in Fig. 5.12.

> Common mistake: Forgetting to mention the boiling point difference requirement.

### Question 9

*3 marks · Short answer*

Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?

**Solution**

1. Evaporation is used to recover a solid solute from a solution by removing the solvent as vapour.
2. Crystallization is used to obtain pure solid crystals from a saturated solution by slow cooling.
3. Distillation is used to separate two miscible liquids or to recover both the solvent and the solute from a solution.

**Answer:** Evaporation recovers solute, crystallization recovers pure crystals, and distillation recovers both components.

> Common mistake: Not distinguishing between recovering just the solute versus recovering both components.

### Question 10

*3 marks · Short answer*

Blood is an example of a colloidal mixture. (i) What would happen if blood behaved like a true suspension inside the body? (ii) In a blood sample, identify the dispersed phase and the dispersion medium.

**Part (i)**

1. If blood behaved like a true suspension, the blood cells would settle down due to gravity, preventing the transport of nutrients and oxygen.

Answer (i): Blood cells would settle down, disrupting vital body functions.

**Part (ii)**

1. In a blood sample, the dispersed phase consists of blood cells (red blood cells, white blood cells, platelets) and the dispersion medium is plasma.

Answer (ii): Dispersed phase: blood cells; Dispersion medium: plasma.

**Answer:** If blood were a suspension, blood cells would settle down over time, which would be fatal for the body. In blood, the dispersed phase is the blood cells and the dispersion medium is plasma.

> Common mistake: Identifying plasma as the dispersed phase instead of the dispersion medium.

### Question 11

*3 marks · Short answer*

You are given a mixture of sand, common salt and naphthalene (Fig. 5.25a). The Fig. 5.25b depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques.

**Solution**

1. Step 1: Sublimation to separate naphthalene from the mixture.
2. Step 2: Filtration to separate sand from the remaining salt solution.
3. Step 3: Evaporation or crystallization to separate common salt from water.

**Answer:** The sequence is: 1. Sublimation, 2. Filtration, 3. Evaporation/Crystallization.

> Common mistake: Reversing the order of filtration and sublimation.

### Question 12

*3 marks · Short answer*

Why is distillation an effective method for separating a mixture of water and acetone?

**Solution**

1. Distillation is effective because acetone and water are miscible liquids with a significant difference in their boiling points.
2. Acetone boils at approximately 56 °C, while water boils at 100 °C.
3. This difference allows acetone to vaporise first, which is then condensed and collected separately.

**Answer:** It is effective due to the large difference in boiling points, allowing for the separation and recovery of both liquids.

> Common mistake: Failing to mention that both liquids can be recovered.

### Question 13

*3 marks · Numerical*

Answer the following questions with the help of the data given in Table 5.4.
(i) What mass of potassium nitrate would be needed to prepare its saturated solution in $50\text{ g}$ of water at $40\text{ }^\circ\text{C}$?
(ii) A student makes a saturated solution of potassium chloride in water at $80\text{ }^\circ\text{C}$ and leaves the solution to cool at room temperature ($25\text{ }^\circ\text{C}$). What would she observe as the solution cools? Explain.
(iii) What is the effect of a change in temperature on the solubility of salts? Also, compare the changes in the solubility of the four given salts with increasing temperature from $10\text{ }^\circ\text{C}$ to $80\text{ }^\circ\text{C}$.

**Part (i)**

1. Solubility of potassium nitrate at 40 °C is 62 g per 100 g of water.
2. Mass needed for 50 g of water = (62 g / 100 g) × 50 g = 31 g.

Answer (i): 31 g

**Part (ii)**

1. Solubility of potassium chloride decreases as temperature drops from 80 °C to 25 °C.
2. The excess solute will separate out as crystals.

Answer (ii): Crystals will form

**Part (iii)**

1. Solubility of solid solutes generally increases with temperature.
2. Potassium nitrate shows the largest increase (21 to 167), while sodium chloride shows the smallest increase (36 to 37).

Answer (iii): Solubility increases with temperature; potassium nitrate increases most, sodium chloride least.

**Answer:** 31 g of potassium nitrate is needed; cooling causes crystallization; solubility generally increases with temperature.

> Common mistake: Forgetting that solubility is defined per 100 g of water.

### Question 14

*3 marks · Numerical*

Three students, A, B and C, are preparing sugar solutions for an experiment:
• Student A dissolves $20\text{ g}$ of sugar in $80\text{ g}$ of water.
• Student B dissolves $20\text{ g}$ of sugar in $100\text{ g}$ of water.
• Student C dissolves $30\text{ g}$ of sugar in $80\text{ g}$ of water.
(i) Calculate the mass percentage ($\%\text{ m/m}$) concentration of sugar in each student's solution.
(ii) Whose solution is the most concentrated? Explain why.

**Part (i)**

1. Student A: Mass of solution = 20 g + 80 g = 100 g; % m/m = (20/100) × 100 = 20%.
2. Student B: Mass of solution = 20 g + 100 g = 120 g; % m/m = (20/120) × 100 = 16.67%.
3. Student C: Mass of solution = 30 g + 80 g = 110 g; % m/m = (30/110) × 100 = 27.27%.

Answer (i): A: 20%, B: 16.67%, C: 27.27%

**Part (ii)**

1. Comparing the percentages, 27.27% > 20% > 16.67%.
2. Student C has the highest mass percentage of solute.

Answer (ii): Student C

**Answer:** Student A: 20%, Student B: 16.67%, Student C: 27.27%; Student C's solution is the most concentrated.

> Common mistake: Using the mass of the solvent as the denominator instead of the total mass of the solution.

### Question 15

*3 marks · Short answer*

Examine Fig. 5.26.
(i) Identify the separation technique marked as 'S'.
(ii) Label the apparatus A, B and C.
(iii) Which of the following mixtures can be separated by the technique identified above? Use the data given in Table 5.5. Mixtures: (a) water — acetone (b) water — salt (c) acetone — alcohol (d) sand — salt (e) alcohol — chloroform (f) alcohol — benzene

**Part (i)**

1. The apparatus shown is used for separating miscible liquids or a liquid from a solution, which is distillation.

Answer (i): Distillation

**Part (ii)**

1. A is the distillation flask containing the mixture.
2. B is the water condenser for cooling vapours.
3. C is the receiver flask (conical flask) for collecting the distillate.

Answer (ii): A: Distillation flask, B: Condenser, C: Receiver flask

**Part (iii)**

1. Distillation requires a boiling point difference of at least 25 °C.
2. Water (100 °C) and acetone (56 °C) have a difference of 44 °C, so they can be separated.
3. Water and salt can be separated as salt is a non-volatile solid.

Answer (iii): (a) and (b)

**Answer:** Technique S is distillation; A is distillation flask, B is condenser, C is receiver flask; mixtures (a) and (b) can be separated.

> Common mistake: Assuming all miscible liquids can be separated by simple distillation regardless of boiling point difference.

## Frequently asked questions

### How many total questions are there in NCERT Solutions for Class 9 Science Chapter 5?

This chapter includes a total of 31 questions across various sections like 'Think It Over', activities, 'Pause and Ponder', and 'Revise, Reflect, Refine'. SwaVid's free PDF and step-by-step solutions are on this page only to help you practice every single question effectively.

### Which topics are covered in the SwaVid solutions for this chapter?

The solutions cover essential topics based on the new NCERT book for the 2026-27 session, including classification of mixtures, solutions, suspensions, colloids, the Tyndall effect, and separation techniques like distillation and crystallization. You can access SwaVid's free PDF and step-by-step solutions on this page only for a complete understanding of these concepts.

### What are the hardest question types in this chapter and how should I approach them?

Numerical problems based on percentage concentration, mass by mass percentage, and volume by volume percentage are often considered the trickiest. To approach them, clearly write down the given mass or volume of solute and solution, apply the correct formula, and double-check your calculations.

### How can I write answers to score full marks in Class 9 Science exams?

To score full marks, structure your answers logically by defining key terms, stating relevant scientific principles, and giving examples where necessary. For numericals and separation techniques, write each step clearly and mention standard units. SwaVid's free PDF and step-by-step solutions on this page only provide ideal answer formats for your practice.

### Is the PDF for Class 9 Science Chapter 5 solutions available for free download?

Yes, complete solutions aligned with the latest NCF 2023 curriculum for the 2026-27 session are available here. SwaVid's free PDF and step-by-step solutions are on this page only to support your exam preparation without any hassle.

## Related pages

- [Class 9 Science chapters](https://www.swavid.com/science/class/9)

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